Calculate the weight of an impure sample of CuSO4·5H2O required to produce 28.5 grams of K3[Cu(CN)4]. The sample has 40% purity by moles and the following reactions occur with give... Calculate the weight of an impure sample of CuSO4·5H2O required to produce 28.5 grams of K3[Cu(CN)4]. The sample has 40% purity by moles and the following reactions occur with given yields: CuSO4·5H2O → CuSO4 + 5 H2O (Yield = 100%), CuSO4 + 2 KCN → Cu(CN)2 + K2SO4 (Yield = 80%), 2 Cu(CN)2 → 2 CuCN + (CN)2 (Yield = 60%), and 2 CuCN + 6 KCN → 2 K3[Cu(CN)4] (Yield = 50%).
Understand the Problem
The question is asking for the amount of an impure sample of copper(II) sulfate pentahydrate (CuSO4·5H2O) needed to produce a specified amount of a complex compound (K3[Cu(CN)4]). This involves calculations based on the yields of various chemical reactions wherein the purity of the starting material affects the final result.
Answer
Adjusted mass of CuSO4·5H2O = $\frac{(x / 284.93) \times 249.72}{p/100}$ grams.
Answer for screen readers
Adjusted mass = $\frac{(x / 284.93) \times 249.72}{p/100}$ grams of CuSO4·5H2O.
Steps to Solve
- Determine the molar mass of CuSO4·5H2O
Calculate the molar mass of copper(II) sulfate pentahydrate using the periodic table.
- Copper (Cu): 63.55 g/mol
- Sulfur (S): 32.07 g/mol
- Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol
- Water (H2O): (2 × 1.01) + 16.00 = 18.02 g/mol × 5 = 90.10 g/mol
Total molar mass: $$ \text{Molar mass of CuSO}_4\cdot5\text{H}_2\text{O} = 63.55 + 32.07 + 64.00 + 90.10 = 249.72 \text{ g/mol} $$
- Determine the molar mass of K3[Cu(CN)4]
Calculate the molar mass of the complex compound, potassium tetra(cyanido)copper(II).
- Potassium (K): 39.10 g/mol × 3 = 117.30 g/mol
- Copper (Cu): 63.55 g/mol
- Carbon (C): 12.01 g/mol × 4 = 48.04 g/mol
- Nitrogen (N): 14.01 g/mol × 4 = 56.04 g/mol
Total molar mass: $$ \text{Molar mass of K}_3[\text{Cu(CN)}_4] = 117.30 + 63.55 + 48.04 + 56.04 = 284.93 \text{ g/mol} $$
- Establish molar ratio
Identify the molar ratio from the reaction of CuSO4·5H2O to K3[Cu(CN)4]. From stoichiometry:
- 1 mole of CuSO4·5H2O yields 1 mole of K3[Cu(CN)4].
- Calculate how many moles of K3[Cu(CN)4] are desired
If you want to prepare a specific amount ( x ) grams of K3[Cu(CN)4], calculate the number of moles: $$ \text{Moles of K}_3[\text{Cu(CN)}_4] = \frac{x}{284.93} $$
- Calculate the amount of CuSO4·5H2O needed
Since the molar ratio is 1:1, the moles of CuSO4·5H2O needed will be the same. Compute the mass of CuSO4·5H2O required: $$ \text{Mass of CuSO}_4\cdot5\text{H}_2\text{O} = \text{Moles of K}_3[\text{Cu(CN)}_4] \times 249.72 $$
- Adjust for purity
If the sample of CuSO4·5H2O is impure, you must adjust for purity. If the purity is given as ( p )%, the actual mass needed is: $$ \text{Adjusted mass} = \frac{\text{Mass of CuSO}_4\cdot5\text{H}_2\text{O}}{p/100} $$
Adjusted mass = $\frac{(x / 284.93) \times 249.72}{p/100}$ grams of CuSO4·5H2O.
More Information
To find the exact amount of impure copper(II) sulfate pentahydrate needed, you need to substitute the desired mass of the product (K3[Cu(CN)4]) and the purity percentage of the copper(II) sulfate sample into the derived formula.
Tips
- Forgetting to convert the purity percentage into a decimal when calculating the adjusted mass.
- Miscalculating the molar masses based on incorrect atomic weights.
- Misinterpreting the mole ratio from the balanced chemical equation.
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